Question: $f(x, y) = -x^3+2y^2+x$ What are all the critical points of $f$ ? Choose 2 answers: Choose 2 answers: (Choice A) A $\left( \dfrac{\sqrt{3}}{3}, 0 \right)$ (Choice B) B $\left( -\dfrac{\sqrt{3}}{3}, 0 \right)$ (Choice C) C $\left( -1, 0 \right)$ (Choice D) D $\left( 1, 0 \right)$
Answer: A critical point of a scalar field $f$ is where $\nabla f = \bold{0}$. [What's that bolded 0?] Let's find the gradient of $f$ ! $\nabla f = \begin{bmatrix} -3x^2 + 1 \\ \\ 4y \end{bmatrix}$ We want each component of the gradient to equal zero, so we want to solve the system of equations below. $\begin{cases} -3x^2 + 1 = 0 \\ \\ 4y = 0 \end{cases}$ The second equation forces $y = 0$. We can rewrite the first equation to say: $x^2 = \dfrac{1}{3}$ Solving, we get $x = \pm \dfrac{\sqrt{3}}{3}$. Therefore, $f$ has critical points at: $\begin{aligned} &\left( \dfrac{\sqrt{3}}{3}, 0 \right) \\ \\ &\left( -\dfrac{\sqrt{3}}{3}, 0 \right) \end{aligned}$